Past exam of the options I course of the University of Oxford / 2011 / P / 2

Set $x=\cos \theta$ with $0\le \theta \le \pi$. The cosine addition identity giveshence $T_0=1$, $T_1=x$, and $T_{n+1}=2x{T}_n-T_{n-1}$. By mathematical induction, $T_n$ is a polynomial of degree $n$: the initial cases hold, and the leading term of $2x{T}_n$ has degree $n+1$ while the subtraction has lower degree. Its leading coefficient is $2^{n-1}$ for $n\ge 1$.

The same change of variables gives the weighted integral

Since $T_2=2x^2-1$, take $x_1=-1/\sqrt{2}$, $x_2=1/\sqrt{2}$. Exactness for $T_0$ and $T_1$ requires $A+B=\pi$ and $(-A+B)/\sqrt{2}=0$. ThusFor $T_2$, both node values and its weighted integral vanish. For $T_3=4x^3-3x$, oddness and symmetric nodes likewise give zero on both sides. Therefore $R(T_j)=0$ for $j=0,1,2,3$. These four Chebyshev polynomials have successive degrees zero through three and form a basis of the degree-at-most-three polynomial space. By linearity, the rule is exact on that whole space: this is two-node Gauss–Chebyshev quadrature.

At either quadrature node, $2x_j^2=1$, so $u(x_j)=\arctan 1=\pi /4$. Thus the Gaussian quadrature approximation isThis is the requested quadrature value; the integrand is not a cubic polynomial, so the degree-three exactness established in part (b) does not make this the exact integral.